1
$\begingroup$

Let $$U \mapsto U \otimes U^* \otimes U \otimes U^*$$ be a unitary representation of the unitary group $U(n)$ acting on the vector space $V$ (where $U^*$ is the complex conjugate of $U$). We can decompose $V$ into invariant subspaces in which the representation is irreducible. Let $P_i$'s be projectors onto invariant subspaces. I want to find these projectors explicitly and use them in a numerical calculation.

Is there any way to do this? like an algorithm for generating them numerically?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
1
$\begingroup$

One way to do this is to solve the Clebsch-Gordan problem. This paper discusses an algorithm for this on the unitary group.

A similar question was asked earlier (see here). I posted an answer there that might also be useful.

Improve this answer
$\endgroup$
5
  • $\begingroup$ I know it has been a long time since I asked the question, but I still have some problems. One is that, when I try to decompose $U\otimes U^*$ with LiE, I get $(0^{n-1})\oplus (1,0^{n-3},1)$ instead of the trivial and the adjoint representation as you mentioned in your answer, and I don't understand why? $\endgroup$
    – Atnap
    Commented Jul 6, 2015 at 19:41
  • $\begingroup$ Atnap: Thanks for your question. I have a typo in that answer. $(1,0^{n-2})$ is the label of the fundamental representation $U$ and $(0^{n-2},1)$ is its conjugate $U^\ast$. The label of the adjoint is $(1,0^{n-3},1)$ not $(1,0^{n-2})$. So $U\otimes U^\ast$ gives you what you got - the trivial and the adjoint. I checked the rest of that answer. There should be no more typos (I hope). $\endgroup$
    – Hari Krovi
    Commented Jul 6, 2015 at 21:45
  • $\begingroup$ Thank you for your response. I also have a question about calculation of Clebsch-Gordan coefficients. I use an online implementation (homepages.physik.uni-muenchen.de/~Arne.Alex/webcleb/) of the algorithm (arxiv.org/abs/1009.0437) you mentioned. For $U\otimes U^*$ with $n=3$, the Dynkin labels are $(1,0)\otimes(0,1)$, so the associated GT labels used in the program are $(1,0,0)\otimes(1,1,0)$ (a GT label specifies the number of boxes in the rows of the corresponding Young diagram), but the coefficients obtained by using these labels do not match the expected trivial and adjoint subspaces. $\endgroup$
    – Atnap
    Commented Jul 7, 2015 at 16:39
  • $\begingroup$ and for example the trivial subspace is span of $\frac{1}{3}(|1,3\rangle-|2,2\rangle+|3,1\rangle)$. However, for $U \otimes U$, the program specifies the symmetric and anti-symmetric subspaces correctly. Any suggestion? $\endgroup$
    – Atnap
    Commented Jul 7, 2015 at 16:40
  • 1
    $\begingroup$ The trivial subspace must be the span of $\frac{1}{\sqrt{3}}(|1,1^\ast\rangle+|2,2^\ast\rangle+|3,3^\ast\rangle)$. For the conjugate $|1^\ast\rangle$ is $|3\rangle$ etc. That explains everything except the minus sign for $|2,2\rangle$ (also I assume the normalization is $\frac{1}{\sqrt{3}}$, not $\frac{1}{3}$). Perhaps we are misinterpreting something in the output of the code. You could try emailing the author. If I figure it out, I'll let you know. $\endgroup$
    – Hari Krovi
    Commented Jul 8, 2015 at 15:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .